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From: joes <noreply@example.org>
Newsgroups: sci.math
Subject: Re: Incompleteness of Cantor's enumeration of the rational numbers
Date: Sun, 24 Nov 2024 14:40:55 -0000 (UTC)
Organization: i2pn2 (i2pn.org)
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Am Sun, 24 Nov 2024 11:31:00 +0100 schrieb WM:
> On 24.11.2024 03:22, Richard Damon wrote:
>> On 11/23/24 4:11 PM, WM wrote:
> 
>>>>> Cover the unit intervals of prime numbers by red hats. Then shift
>>>>> the red hats so that all unit intervals of the positive real axis
>>>>> get red hats.
>>>> And you can, as the red hat on the number 2, can be moved to the
>>>> number 1 the red hat on the number 3, can be moved to the number 2
>>>> the red hat on the number 5, can be moved to the number 3
>>>
> It fails in every step to cover the interval (0, n] with hats taken from
> this interval.
Nobody cares about a nonbijection from N to N. We are interested in a
bijection from N to P, or {0, 1, ..., n} to {p0, p1, ..., p_n}

>>> Yes, for every n that belongs to a tiny initial segment. 
>> No, for EVERY n. Show one that it doesn't work for!
> The complete covering fails in every interval (0, n] with hats taken
> from this interval.
Think about it the other way around: when we take the primes and
number them, we are never done, because there are inf.many primes.
Therefore we need the set of all naturals, N, to count them. Then
there can't be fewer primes than naturals, since the cardinality
of N is the smallest infinity.

>>>> so all the numbers get covered.
>>> No.
>> WHich one doesn't.
> Almost all. For every interval (0, n]
> the relative covering is 1/10, independent of how the hats are shifted.
As above, nobody is arguing that the primes are somehow more infinite.

> This cannot be remedied in the infinite limit because outside of all
> finite intervals (0, n] there are no further hats available.
Yes it can, if you start out with a proper finite bijection.

> On the other hand, we cannot find a first n that cannot be covered by a
> hat. This dilemma cannot be resolved by negating one of the two facts.
> It can only be solved by dark numbers.
Or getting over your wrong intuition.

-- 
Am Sat, 20 Jul 2024 12:35:31 +0000 schrieb WM in sci.math:
It is not guaranteed that n+1 exists for every n.